www.rtmj.net > 已知等差数列{An}公差不为0,且A3=5,A1,A2,A5成等比数列.(1)求数列{An}的通项公...

已知等差数列{An}公差不为0,且A3=5,A1,A2,A5成等比数列.(1)求数列{An}的通项公...

(1)在等差数列{an}中,设其公差为d,(d≠0),∵a1a5=a22,a3=5,∴(a3-2d)(a3+2d)=a22,即(5-2d)(5+2d)=(5-d)2,…2分 化简得5d2-10d=0,∴d=2…4分 ∴an=a3+(n-3)d=5+2(n-3)=2n-1…7分 (2)∵b1+2b2+22b3+…+2n-1bn=an,① ∴b1+2b2

1)因为an为等差数列所以a1=5-2d a2=5-d a5=5+2d又a1,a2,a5成等比数列所以(a2)^2=a1*a5 既(5-d)^2=(5-2d)*(5+2d) 又d≠0 解得d=2 则a1=1an=a1+(n-1)d=1+(n-1)*2=2n-12)b1+2*b2+2^2*b3++2^n-1*bn=anb1+2*b2+2^2

(I)设等差数列的公差为d由题意可得, a1a5=a22 a3=5 ∴ a1(a1+4d)=(a1+d)2 a1+2d=5 解可得, a1=1 d=2 ∴Sn=na1+ n(n?1)d 2 =n+n(n-1)=n2(II)∵b1+2b2+4b3+…+2n-1bn=an,∴b1+2b2+4b3+…+2n-1bn=an,

题目是不是打错了由{an}为等差数列,得a1=a3-2d=5-2d,a2=a3-d=5-d由a1,a2,a3成等比数列,得(5-2d)/(5-d)=(5-d)/5解得d=0 与题干中的“公差不为0”矛盾

a1a2a5成等比数列a2=a1*a5(a1+d)=a1(a1+4d)解得d=2a1,d=0(舍去)∵a3=5∴a3=a1+2d=5a1=5a1=1d=2等差数列an=a1+(n-1)d=1+2n-2=2n-1如果您认可我的回答,请点击“采纳为满意答案”,祝学习进步!手机提问的朋友在客户端右上角评价点【满意】即可

数列a1 a2 a5 等比数列则有a2*a2=a1*a5 a3-2d=a1 a3+2d=a5 a3-d=a2带入得到d=2b1+2b2+4b3+2^(n-1)bn=an (1)b1+2b2+4b3+2(n-3)b n-1=a n-1 (2)(1)-(2)得到2^(n-1)bn=an - a n-1=2解得bn=2/2^(n-1)=4/2^nb n-1=4/2^(n-1)bn/b n-1=1/2 公比q=1/2n=1带入得到b1=2 首项为2即(bn)为等比数列Tn=4-2^(3-n)第2问比大小你自己搞定了没时间了!

数学知识都还给老师了哈哈,回答你第一问.a1+a2=2a1+d=1a2,a3,a5成等比数劫,a5/a3=a3/a2,即(a1+4d)/(a1+2d)=(a1+2d)/(a1+d),化简这个等式,得到a1d=0,因为d不等于0,所以a1=0.2a1+d=1,得知d=1所以数列{an}的通项公式为:an=a1+(n-1)d=n-1第二问不会啦哈哈,望采纳

(1)∵a1,a2,a5成等比数列,a3=5∴a22=a1a3∴(5-d)2=(5-2d)(5+2d)∵d≠0∴d=2∴an=a3+(n-3)d=5+2(n-3)=2n-1(2)由(1)可得,Sn= 1+2n?1 2 *n=n2∴bn=2n? Sn =n?2n∴Tn=1?2+2?22+3?23+…+n?2n∴2Tn=1?22+2?23+…+(n-1)?2n+n?2n+1两式相减可得,-Tn=2+22+23+…+2n-n?2n+1= 2(1?2n) 1?2 ?n?2n+1=2n+1-2-n?2n+1∴Tn=(n?2)?2n+1+2

设公差为d因为a1,a2,a3成等比数列所以(a2)=a1a3即(a1+d)=a1(a1+2d)(a1)+2a1d+d=(a1)+2a1d得d=0,与题目的公差d不为零矛盾,所以题目有问题

解:(1)设公差为d,d≠0a2,a3,a5成等比数列,则a3=a2a5(a1+2d)=(a1+d)(a1+4d)整理,得a1d=0d≠0,因此只有a1=0a1+a2=1a2=1-a1=1-0=1d=a2-a1=1-0=1数列{an}是以0为首项,1为公差的等差数列an=0+1*(n-1)=n-1数列{an}的通项公式为an=n-1(2)bn=an+2^(an)=(n-1)+2^(n-1)Tn=b1+b2++bn=0+1+1+2++(n-1)+2^(n-1)=[0+1++(n-1)]+[1+2++2^(n-1)]=n(n-1)/2 +1*(2^n -1)/(2-1)=n(n-1)/2 +2^n -1

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